I did the case for a $(0, 1)$ tensor $A = A_i$, it's
$$ (\Delta A)_i = \Delta (A_i) - g^{jk}\partial _k \Gamma_{ij}^l A_l - 2 g^{jk}\Gamma_{ij}^l A_{l,k} +g^{jk}\Gamma_{ik}^l \Gamma_{jl}^m A_m + g^{jk} \Gamma_{jk}^l \Gamma_{il}^m A_m$$
In general there is no hope that $(\Delta A)_i = \Delta (A_i)$, since the equations should be all coupled and the term $g^{jk}\partial _k \Gamma_{ij}^l A_l$ involves the curvature. If your manifold is something explicit, might be you can simplify the term though.
I've seen in the literature the notation $C$ with some additional specifications for the contraction maps of all sorts, but the amount of decorations on the symbol $C$ varied depending on the context. See, e.g., A.Gray, Tubes, p.56, where these maps are used in the case of somewhat special tensors, and therefore the notation is simpler.
In general, there is a whole family of uniquely defined maps
$$
C^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \otimes^{r-1}_{s-1} V
$$
which are collectively called tensor contractions ($1 \le p \le r, 1 \le q \le s$).
These maps are uniquely characterized by making the following diagrams commutative:
$$
\require{AMScd}
\begin{CD}
\times^{r}_{s} V @> {P^{(r,s)}_{p,q}} >> \times^{r-1}_{s-1} V\\
@V{\otimes^{r}_{s}}VV @VV{\otimes^{r-1}_{s-1}}V \\
\otimes^{r}_{s} V @>{C^{(r,s)}_{p,q}}>> \otimes^{r-1}_{s-1} V
\end{CD}
$$
Explanations are in order.
Recall that the tensor products $\otimes^{r}_{s} V$ are equipped with the universal maps
$$
\otimes^{r}_{s} \colon \times^{r}_{s} V \to \otimes^{r}_{s} V
$$
where $\times^{r}_{s} V := ( \times^r V) \times (\times^s V^*)$.
Besides that, there is a canonical pairing $P$ between a vector space $V$ and its dual:
$$
P \colon V \times V^* \to \mathbb{R} \colon (v, \omega) \mapsto \omega(v)
$$
Notice that map $P$ is bilinear and can be extended to a family of multilinear maps
$$
P^{(r,s)}_{p,q} \colon \times^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
by the formula:
$$
P^{(r,s)}_{p,q} (v_1, \dots, v_p, \dots, v_r, \omega_1, \dots, \omega_q, \dots, \omega_s) = \omega_q (v_p) (v_1, \dots, \widehat{v_p}, \dots, v_r, \omega_1, \dots, \widehat{\omega_q}, \dots, \omega_s)
$$
where a hat means omission.
Since maps $P^{(r,s)}_{p,q}$ are multilinear, the universal property of the maps $\otimes^{r}_{s}$ implies that there are uniquely defined maps
$$
\tilde{P}^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
and then the maps $C^{(r,s)}_{p,q}$ are given by
$$
C^{(r,s)}_{p,q} := \otimes^{r-1}_{s-1} \circ \tilde{P}^{(r,s)}_{p,q}
$$
Best Answer
Posting an answer so as to mark this query as resolved. The formula on mathworld is just the trace of a double covariant derivative, the second of which has been raised using the metric. I.e,
$$(\Delta\mathbf{T})^{i_1\dots i_r}{}_{j_1\dots j_s}=T^{i_1\dots i_r}{}_{j_1\dots j_s~;k}{}^{;k}=g^{kl}T^{i_1\dots i_r}{}_{j_1\dots j_s~;k;l}$$ $$=g^{kl}(\nabla(\nabla\mathbf{T}))^{i_1\dots i_r}{}_{j_1\dots j_s~kl}$$
But better still is to just write $$\Delta=\nabla^i\nabla_i$$